Solving DDEs in MATLAB

More documents

Recommendations

Info

444 L.F. Shampine, S. Thompson / Applied Numerical Mathematics 37 (2001) 441–458 is an explicit recipe for the stage and the formulas are explicit. The function S(x) is the initial history for x a. After taking the step to xn+1, we use the continuous extension to define S(x) on [xn,xn+1] as S(xn + σhn) = yn+σ . We are then ready to take another step. This suffices for proving convergence as the maximum step size tends to zero, but we must take up the other situation because in practice, we may very well want to use a step size larger than the smallest delay. When hn >τj for some j, the “history” term S(x) is evaluated in the span of the current step and the formulas are defined implicitly. In this situation we evaluate the formulas with simple iteration. On reaching xn, wehavedefinedS(x) for x xn. We extend its definition somehow to (xn,xn + hn] and call the resulting function S (0) (x). A typical stage of simple iteration begins with the approximate solution S (m) (x). The next iterate is computed with the explicit formula S (m+1) (xn + σhn) = yn + hnΦ xn,yn,σ; S (m) (x) . Here and elsewhere it is convenient to show the dependence on the history by means of another argument in the increment function. In the next section we show that if the step size is small enough, this is a contraction and S(x) is well-defined for x xn+1. In dde23 we predict S (0) (x) to be the constant y0 for the first step. We do not attempt to predict more accurately then because the solution is not smooth at a and we do not yet know a step size appropriate to the scale of the problem. Indeed, reducing the step size as needed to obtain convergence of simple iteration is a valuable tactic for finding an initial step size that is on scale, cf. [16]. After the first step, we use the continuous extension of the preceding step as S (0) (x) for the current step. This prediction has an appropriate order of accuracy and we can even assess the accuracy quantitatively using (2). Remarkably, the truncation error has a local maximum at xn+1 so that extrapolation actually provides a more accurate solution for some distance. Specifically, the ratio of the error of yn+σ to the error of yn+1 is no larger than 1 for σ up to about 1.32. Much more important in practice is the accuracy of the prediction in the span of a subsequent step of the same size or rather larger. Unfortunately, the ratio grows rapidly as σ increases. Still, it is no bigger than 13 for 1 σ 2, so the prediction is quite good when the step size is changing slowly. The step size adjustment algorithms of dde23 are those of ode23 augmented to deal with implicit formulas. The convergence test for simple iteration is that y (m+1) n+1 − y(m) n+1 is no more than one tenth the accuracy required of yn+1. Because each iteration costs as much as an explicit step, any proposed hn with τ
L.F. Shampine, S. Thompson / Applied Numerical Mathematics 37 (2001) 441–458 445 This theorem shows how our method works. A more general convergence result may be found in [21]. Before proving this result, we begin our investigation of convergence by recalling the comments earlier about how discontinuities arise and propagate for DDEs. We know in advance all the points where the solution might not be smooth enough for the usual orders of the formulas to hold. Because we consider one-step methods, if we take these points to be mesh points, the analysis is the same as if the solution were smooth throughout the interval. Accordingly, we consider a sequence of meshes {xn} which include all these points. Just how this is accomplished in practice is described in Section 5. Between these points we assume that f and y(x) are smooth enough for the formulas to have their usual orders. We also assume that f satisfies a Lipschitz condition with respect to all the dependent variables. For the sake of simplicity, we suppose that there is only one delay and take the Lipschitz condition in the form f(x,˜y, ˜z) − f(x,y,z) L max ˜y − y, ˜z − z . The Runge–Kutta formulas extended to DDEs involve a history term that we write generically in the increment function as g(x). We require two lemmas about how the increment functions depend on their arguments. Lemma 1. There is a constant L such that for 0 σ 1, Φ xn, ˜yn,σ; g(x) − Φ xn,yn,σ; g(x) L˜yn − yn. (4) Proof. In a step of size hn from (xn, ˜yn), the intermediate quantities ˜yni are defined by i−1 ˜yni =˜yn + hn aij f xnj , ˜ynj ,g(xnj − τ) j=1 and the yni are defined similarly in a step from (xn,yn). Using the Lipschitz condition on f ,itis straightforward to show that for each i, there is a constant Li such that i−1 ˜yni − yni 1 + HL Lj|aij | ˜yn − yn=Li˜yn − yn. j=1 With this result it follows easily that (4) holds with L = L s i=1 Li max(|bi(σ )|). ✷ The second lemma we need is a bound on the effect of using different history terms. Lemma 2. Let ∆ be a bound on G(x) − g(x) for all x xn+1. If the maximum step size H is small enough that i−1 |aij | 1 (5) HLmax i j=1 then there is a constant Γ such that Φ xn,yn,σ; G(x) − Φ xn,yn,σ; g(x) Γ∆. (6) Proof. Let the intermediate quantities of the two formulas be i−1 rni = yn + hn aij f xnj ,rnj ,G(xnj − τ) j=1
Page 1 and 2: Applied Numerical Mathematics 37 (2
Page 3: L.F. Shampine, S. Thompson / Applie
Page 7 and 8: L.F. Shampine, S. Thompson / Applie

Solving DDEs in MATLAB

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?